JEE Main · 2020 · Shift-ImediumBIO-054

Consider the following reactions: (i) Glucose + ROH dry HCl Acetal [(CH3CO)2O]x eq. of Acetyl derivative (ii) Glucose…

Biomolecules · Class 12 · JEE Main Previous Year Question

Question

Consider the following reactions:

(i) Glucose+\ceROH→dry HClAcetal→\ce(CH3CO)2Ox eq. ofAcetyl derivative\text{Glucose} + \ce{ROH} \xrightarrow{\text{dry HCl}} \text{Acetal} \xrightarrow[\ce{(CH3CO)2O}]{\text{x eq. of}} \text{Acetyl derivative}

(ii) Glucose→\ceNi/H2A→\ce(CH3CO)2Oy eq. ofAcetyl derivative\text{Glucose} \xrightarrow{\ce{Ni/H2}} \text{A} \xrightarrow[\ce{(CH3CO)2O}]{\text{y eq. of}} \text{Acetyl derivative}

(iii) Glucose→\ce(CH3CO)2Oz eq. ofAcetyl derivative\text{Glucose} \xrightarrow[\ce{(CH3CO)2O}]{\text{z eq. of}} \text{Acetyl derivative}

'x', 'y' and 'z' in these reactions are respectively:

Options
  1. a

    4, 5 and 5

  2. b

    4, 6 and 5

    ✓
  3. c

    5, 4 and 5

  4. d

    5, 6 and 5

Correct Answerb

4, 6 and 5

Detailed Solution

Step 1: Analyzing Reaction (i) — Glucoside formation blocks the anomeric hydroxyl (−\ceOH-\ce{OH}) at C1\text{C1}. The remaining four −\ceOH-\ce{OH} groups react with acetic anhydride. Hence, x = 4.

Step 2: Analyzing Reaction (ii) — Reduction with \ceNi/H2\ce{Ni/H2} converts the aldehyde into a primary alcohol, yielding sorbitol (C6H14O6C_6H_{14}O_6), which has six −\ceOH-\ce{OH} groups. Hence, y = 6.

Step 3: Analyzing Reaction (iii) — Normal glucose directly forms a pentaacetate using its five available −\ceOH-\ce{OH} groups. Hence, z = 5. The sequence is 4, 6, 5.

Key Point: The number of equivalents used indicates the total number of free hydroxyl groups in the molecule.

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